# Solving for initial current for an RLC circuit with step voltage source

I have a problem where I am asked to find $$\i_L(t)\$$ and $$\v_c(t)\$$ for a parallel RLC circuit, with a step voltage source.

The circuit diagram is this below, where the voltage source is $$\5 u(t)\$$, or $$\V=0, t<0\$$ and $$\V=5, t\ge0\$$.:

So far, I found $$\\alpha=\dfrac{1}{2RC}=3.33 *10^5\$$ rad/s, and $$\\omega_o=\dfrac{1}{\sqrt{LC}}=3.16*10^5\$$ rad/s.

Since $$\\alpha^2 > \omega_0^2\$$, it is overdamped, and thus the general solution of the DE has the form $$\i_L(t)=Ae^{s_1t}+Be^{s_2t}+i_L(\infty)\$$.

The roots of the characteristic equation are, $$\S_1\$$ and $$\S_2\$$ are $$\ \{-2.279*10^5, -4.387*10^5\} \$$.

It is known that:

$$\ i_L(0^-) = 0 \$$ A, since there is no source at that time.

$$\ i_L(\infty) = 0.033 \$$ A, since capacitor would be fully charged (no current flows), and the inductor would essentially act as a short.

The part I'm having trouble is determining what $$\i_L(0^+)\$$ would be. In this circuit, how would the inductor current be determined immediately after $$\t = 0\$$?

## 2 Answers

The current through an (ideal) inductor is continuous, just as the electric potential difference over a capacitor. So if the current $$\ i_L \$$ at $$\t=0-\$$ is zero, then so is the current at $$\t=0+\$$, immediately after the voltage source stepped to $$\+5 V\$$.

If you want a full solution of capacitor voltage and inductor current, you'd have to solve the differential equation for e.g. the capacitor voltage $$\ u_C\$$. $$u_S(t) = 5 * UnitStep$$ $$i_R(t) = (u_S(t) - u_C(t))/R$$ $$i_L(t) = \frac{1}{L}\int_0^t u_C(\tau) d\tau$$ $$i_C(t) = C \frac{du_C}{dt}$$

In which the subscripts are quite self-explaining, except $$\u_S(t)\$$, which is the source voltage.

Now, omitting the time where possible, we can write $$u_c = u_S - R(i_L + i_C) = u_S - R\left(\frac{1}{L}\int u_C dt +C\frac{du_C}{dt}\right)$$ After differentiating the whole expression we get, and we write u for u_C, $$u'= -\frac{R}{L}u-RCu''$$ or $$RCu''+u'+\frac{R}{L}u = 0$$ Substituting the values for $$\R\$$, $$\L\$$ and $$\C\$$ gives $$150*10^{-8}u''+u'+ 150*10^3u = 0$$ Multiplying with $$\2*10^6\$$ gives us $$3u''(t) + 2 * 10^6 u'(t) + 3 * 10^{11} u(t) = 0$$

Some boundary conditions of this problem are $$\u(0)\$$ and $$\u'(0)\$$. Of course $$\u(0)=0\$$ as at the start the capacitor is empty. $$\u'(0)\$$ can be found by realizing ourselves that $$\i_L(0) = 0\$$ because the current in any self inductance is continuous and therefore all the current at $$\t=0\$$ will flow into the capacitor. This current is caused by the voltage drop of $$\5 V\$$ over the resistors when the step is enacted and C has zero charge hence zero voltage. So this current is $$\ i_C(0+) = 5/150 A \$$ which causes a rate of change of the capacitor voltage $$\ u'_C(0+) = C*i_C(0+) = \frac{5}{150}*10^{-8} V/s\$$. Therefore the boundary conditions are $$u(0) = 0$$ $$u'(0) = \frac{1}{3} * 10^7$$

$$u_C(t) = 5 * \sqrt 10 * e^{-10^5 (10 + \sqrt 10) t/3} (e^{2*10^5 \sqrt(10) t/3} - 1)$$ and it plots like this.

Now this deals with the voltage $$\u_C(t)\$$ over the capacitor and your question was regarding the inductor current which is $$i_L(t) = \frac{1}{L} \int_0^t u(\tau) d\tau$$ and with $$\a=\sqrt 10\$$, $$\b=100000/3\$$ (and $$\a^2=10\$$) this becomes $$i_L(t) = \frac{1 - e^{-a^2 b t} \left(cosh(a b t) + a sinh(a b t)\right)}{b(1 - 1/a^2)}$$

Of course, when $$\t=0\$$, the exponent becomes $$\e^{-0}=1\$$, $$\sinh(0)=0\$$ and $$\cosh(0)=1\$$, hence the whole expression becomes $$i_L(0+) = \frac{1-1}{b(1-1/a^2)}=0$$

The part I'm having trouble is determining what iL(0+) would be. In this circuit, how would the inductor current be determined immediately after t=0?

As with any inductor, the current cannot change instantly so, the current immediately after t = 0 is zero.